Problem in counting 2 and 7

[DSchles]

If the T is worth 1 then the ace is worth 1.2

You can't make a blanket statement like that without knowing the rules, and, in any event, for almost any set of rules, it isn't true. Table D17, p. 522 of BJA3 gives EORs under all different rules sets. For example, for S17, DAS, the ace is worth 1.13 times the ten. But for H17, DAS, the TEN is worth 1.065 times the ace! And, if surrender is offered in a S17 DAS game, the ace is worth 1.027 times the ten.

In short, the ace is NOT worth 1.2 times more under any circumstances, and sometimes it is worth much less, particularly in H17 games.

And wouldn't it be nice, since the numbers are staring everyone in the face, that we didn't make a 98-post thread on something that merits NO discussion whatsoever??

Don

[Phoebe]

Hence the interest of using MGP's BJCA (apart from reading BJA3)
Free but I feel like I'm the only one using it.


BTW, my sample was with surrender

[Three]

Well if you look at Peter Griffin's TOB p25, it lists the EoR of the ace as 3.1 and the T as 2.6 for a ratio of 1.2:1. Knowing Griffin's work that os probably for SD with rules that are hard to find today.

6 deck EoR's at A to T ratios in bold:
http://www.bjstrat.net/ccTheory.html bjs
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"Considering multiple decks

In developing a counting system it should be applicable to any number of decks so let's next consider 6 decks as an example (using the same rules as above.) When one of any rank is removed from 6 full decks the eor is numerically much less than the eor from a full single deck. This is because removing one card has less effect on a shoe's composition the more decks there are. First we are going to list the actual eors for 6 decks. Following that we are going to normalize them so we can compare single deck eors to 6 deck eors. Normalization is a statistical method for doing comparisons. The formuls used is -

Normalized_EOR = EOR * (52 * (deck_size) - 1) / 52,
where deck_size is number of decks and EOR is eor of 1 card of a given rank

Six Deck EORs (in %)
2 3 4 5 6 7 8 9 T A
CD Basic .06314 .07210 .09384 .11816 .06945 .04479 -.00235 -.03282 -.08223 -.09739 A to T ratio 1.2:1
TD Basic .0627 .0713 .0911 .1139 .0740 .0445 -.0016 -.0308 -.0816 -.0981 A to T ratio 1.2:1
Generic Basic .0623 .0707 .0906 .1135 .0737 .0444 -.0012 -.0306 -.0814 -.0986 A to T ratio 1.2:1

Normalized Single Deck EORs (in %)
2 3 4 5 6 7 8 9 T A
CD Basic .3751 .4276 .5402 .6854 .4064 .2799 .0081 -.1654 -.4932 -.5842 A to T ratio 1.2:1
TD Basic .3794 .4299 .5264 .6634 .4466 .2830 .0224 -.1582 -.5025 -.5831 A to T ratio 1.2:1
Generic Basic .3635 .4037 .5033 .6331 .4243 .2692 .0191 -.1526 -.4693 -.5860 A to T ratio 1.2:1

Normalized Six Deck EORs (in %)
2 3 4 5 6 7 8 9 T A
CD Basic .3776 .4312 .5612 .7067 .4154 .2679 -.0141 -.1962 -.4918 -.5825 A to T ratio 1.2:1
TD Basic .3750 .4264 .5264 .6812 .4426 .2661 -.0096 -.1842 -.4880 -.5867 A to T ratio 1.2:1
Generic Basic .3726 .4228 .5419 .6788 .4408 .2655 -.0072 -.1830 -.4868 -.5897 A to T ratio 1.2:1
The normalized eors for 1 and 6 decks are reasonably similar. Since we have discarded the optimum count approach as being impractical it does not seem unreasonable to apply whatever the end method turns out to be to any number of decks."
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As EoRs vary by rule set, if I kept looking I am sure I would find rules that don't have a 1.2:1 ratio for the A and T. But the ones I can find easily are all saying a 1.2:1 ratio.

Okay. The WoO site uses the bjstrat's Combinatorial Analyzer and get different values than the guys at bjstrat do with their own CA:
https://wizardofodds.com/games/blackjack/appendix/7/

[ZenMaster_Flash]

Dr. Griffin made it clear that his Single Deck values for
E.O.R.'s and for most other values are interpolated for
multiple decks. So, if a value of X is for single deck then
finding the REDUCED value for 6 or 8 decks are found by
moving from 1 deck to 6 or 8 decks by interpolation
using infinite deck values and moving 1/6 or 1/8 of the
way between the two values e.g. if a value for SD is +.88
and the infinite deck value is +0.12 then it follows that
0.88 + 0.12 = 1.00/6 or 8 = 0.166 or 0.125

Am I keyboarding under the influence? Yes! Did I get
this wrong? I hope not. Your feedback is being elicited.

In The Theory of Blackjack [sixth (6th) ed. only] Dr. Griffin
re-calculated the Single Deck Values for adjusting our play
via the True count for Shoe Games.

You can look at this updated matrix and focus on the
"Illustrated 18" to see the massive reduction in the
values moving from SD to Shoe Games in thousands
of 1% Of course Insurance and correct plays with your
hands of 12 vs. 2 through 6 and splitting 10's vs. 4 - 6
are the most crucial adjustments that are available to us.

If, in the past, you did not fathom why the SD and DD games
really require a count with high P.E. [playing efficiency] for
effective Card Counting. If you play DD games, you can use 1/2
of the SD values ~ which is simple enough, eh? Betting Correlation
for all goodcountsrange from 0.97 to 0.99 ~ but P.E. should at
least be > 0.60. Side Counts are required for a Card Counter to
exceed 0.67 Griffin makes it clear that exceeding 0.70 requires
Side Counts and Side Counting the Aces and Sevens in a DD game.
Yes, there are adjustments to your play that can radically improve
your Playing Efficiency, but including the 5's, 8's, etc. is "overkill"
as they generate what amounts to significantly "diminishing returns"
for those seeking [virtually] "perfect play." via The Gordon Count,
or Hi-Opt I or II [with side-counted Aces, 6's, 7's, 8's], the Tarzan Count.

I suggest that all newbies as well as those who are experienced but seek to
gain a deeper understanding you pay very close attention to Dr. Griffin's work.

[DSchles]

My God, you do know how to waste everyone's time.

It doesn't make the slightest difference if you use EORs for 6 decks or 1 deck. The ratios remain virtually identical to the ones I already gave. 1.13 for S17 DAS and a larger EOR for a ten than for an ace for H17 games.

But, you are going to try to turn this into a 98-post thread, aren't you? The EOR for an ace is NOT 1.2 more than for a ten. Not in BJA3, not on WOO, and not in the combinatorial analyzer. Those still reading this abject nonsense can trust all of the above, or you can read Three's erroneous treatise. Your choice.

Don

[kelg21]

Dr. Griffin made it clear that his Single Deck values for
E.O.R.'s and for most other values are interpolated for
multiple decks.

Don S came up with nifty resolution to all this rhetoric called SCORE. Heard of it?

Am I keyboarding under the influence? Yes!

Must YOU?

Reread Don's explanation. It was perfect.

Hard to argue your logic. Create confusion to confound the understanding. Must you? Why?

[Three]

None of it was mine. I quoted Peter Griffin and bjstrat both of which state the ratio is 1.2:1. The following is the only part of my post that was mine.

As EoRs vary by rule set, if I kept looking I am sure I would find rules that don't have a 1.2:1 ratio for the A and T. But the ones I can find easily are all saying a 1.2:1 ratio.

Okay. The WoO site uses the bjstrat's Combinatorial Analyzer and get different values than the guys at bjstrat do with their own CA:
https://wizardofodds.com/games/blackjack/appendix/7/

I linked to the WoO link that agrees with you.

Those still reading this abject nonsense can trust all of the above, or you can read Three's erroneous treatise.

I guess I can't show what these three respected experts say without catching shit for their research. Don, you do realize you weren't disagreeing with me. You were disagreeing with the links from bjstrat and Peter Griffin's "The Theory of Blackjack", while agreeing with my other sited source WoO. I said they probably all use different rule sets to account for the differences. You just said Peter Griffin and bjstrat are full of it. Personally, I don't play H17 BJ so I don't care what the EoRs are for H17 BJ. I play S17, DAS, LS games. If you want to post the EoRs for those rules I would appreciate it. I am starting to feel I can't trust all these respected BJ experts.

[Jack Jackson]

Why not use a custom count that "omits" the 7 to start with for BOTH betting and playing purposes, INSTEAD of having to count it twice?

Something like this looks feasible (2-A) 11222000-2(0)

[Three]

Why not use a custom count that "omits" the 7 to start with, for BOTH betting and playing purposes, instead of having to count it twice?

Something like this looks feasible (2-A) 11222000-2(0)

Counting a card in the main count and also side counting it takes the same effort as what you are recommending. The adjustments account for it already being in the main count. The stats for Hiopt2 main count and your main count favor Hiopt2 in all 3 stats. Hiopt2 would be dropping the ace and reducing the 6 by 1 in the main count:
PE: Hiopt2 .6684, yours .6325
BC: Hiopt2 (without the ace) .9087, yours (without the ace).8974
IC: Hiopt2 .9085, yours .8777

I would recommend side counting one rank first until he gets used to side counting and then adding another rank. You seem to be recommending starting with two side counts as both the ace and 7 are not counted in the main count.

[Three]

Personally, I don't play H17 BJ so I don't care what the EoRs are for H17 BJ. I play S17, DAS, LS games. If you want to post the EoRs for those rules I would appreciate it. I am starting to feel I can't trust all these respected BJ experts.

Never mind. I found them on p522 of BJA3.
S17, NDAS, DOA, SPA1, SPL3: T -.5081, A -.5944. Ratio 1.2:1
S17, DAS, DOA, SPA1, SPL3: T -.5121, A -.5794. Ratio 1.1:1
S17, NDAS, DOA, SPA1, SPL3, LS: T -.5558, A -.5901. Ratio 1.1:1
S17, DAS, DOA, SPA1, SPL3, LS: T -.5599, A -.5751. Ratio 1.0:1

It seems only in NDAS games does the A reach a ratio of 1.2:1 of the T. I don't play those games.

[Dbs6582]

You can't make a blanket statement like that without knowing the rules, and, in any event, for almost any set of rules, it isn't true. Table D17, p. 522 of BJA3 gives EORs under all different rules sets. For example, for S17, DAS, the ace is worth 1.13 times the ten. But for H17, DAS, the TEN is worth 1.065 times the ace! And, if surrender is offered in a S17 DAS game, the ace is worth 1.027 times the ten.

In short, the ace is NOT worth 1.2 times more under any circumstances, and sometimes it is worth much less, particularly in H17 games.

And wouldn't it be nice, since the numbers are staring everyone in the face, that we didn't make a 98-post thread on something that merits NO discussion whatsoever??

Don

I can't find it now, but I know I've read somewhere where if your first card is an ace, your chances of winning that hand were around 50%. If your first card is a 10, your chances of winning were much lower, like in the 10 to 20% range. I'm probably wrong on my percentages but I know they're higher with a player's first card being an ace.

Also, in my local casinos they will have promotional nights where they send me an ace coupon I can use anytime as my first card. I've read where this is a fairly common promotion. A casinos would never send someone a coupon with a 10 card since most people would intuitively know this isn't very valuable, or at least not as valuable as an ace.

It's for these reasons I've always believed that if my count is positive and I know it's due to an excess in aces rather than an excess in 10s my advantage is greater than the count calls for. This is why I thought it'd make sense to bet more in these situations.

[ZenMaster_Flash]

" ... if your first card is an ace, your chances of winning that hand were around 50%.

This is hilarious. It is sillier than some of the dumber mistakes that I have made.

[tater]

For this, there is BRH. https://www.blackjackreview.com/wp/a...s/brh-systems/

Interesting Article. All the talk about side counting the 7? Simply drop the 2 for pitch games. Use the same brain power for Insurance count.

I can't find it now, but I know I've read somewhere where if your first card is an ace, your chances of winning that hand were around 50%. If your first card is a 10, your chances of winning were much lower, like in the 10 to 20% range. I'm probably wrong on my percentages but I know they're higher with a player's first card being an ace..

Silly? If you don't mind, I will start every hand with an Ace.

Am I keyboarding under the influence? Yes! Did I get
this wrong? I hope not. Your feedback is being elicited.

Rule of thumb in handicaping horses. When in doubt, always seek advice from the drunk guy. It's unlikely you will get the winner but guaranteed you'll receive a fountain of misinformation.